How many GRAMS of fluorine are present in 5.35×1022 molecules of boron trifluoride?

I don’t know if I did this right, but I hope it helps.

Molar mass of BF3
M(BF3) = (1x10.8) + (3x18.91)
= 67.53 grams per mol

Molar mass of Boron in BF3
M(B) = 1x10.8 = 10.8 grams per mol

5.35x10^22 molecules of BF3 into mol
n = N/NA

n(BF3) = 5.35x10^22/6.02x10^23
= 8.89x10^44 mol of BF3

Mass of B in BF3
m = n x M
m(b) = 8.89x10^44 x 10.8
= 9.6x10^45g

Answerit curves up and levels out.explanationwe assume that the container contains absolutely no ammonia to begin with as it states that "nitrogen gas and hydrogen gas" react to "produce ammonia."at dynamic equilibrium, the product and reactant concentrations are equal.this means that, starting from a concentration of zero, [nh₃] must curve "up" from zero and then level out to a constant concentration.

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